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CBSE Class 10 Answered

can u solve
Asked by | 05 May, 2009, 10:31: PM
answered-by-expert Expert Answer

 

From the question we gather that the required number will be 17 less than the smallest common multiple (LCM)of 520 and 468.

The L.C.M of the given numbers =product of the numbers /hcf=4680

So we need to find the number which is less than 4680 and less by 17.

So the reqd number is 4680-17=4663

Check

If we add 17 to 4663 ,

 we get,

 4680 ,

 which is exactly divisible by 520 and 468

Answered by | 07 May, 2009, 02:41: PM

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